Linear Dependence of x1, x2 and x3 in R^2

In summary, the vectors x1, x2, and x3 must be linearly dependent because x1 and x2 span R^2 and x3 can be written as a linear combination of these vectors. Additionally, the dimension of span (x1, x2, and x3) is 1, according to the book, but it could potentially be 3 if the vectors are all the same times a constant. The definition of dimension must be applied to determine the correct dimension.
  • #1
2,285
5
x1= column vector (2, 1)
x2= column vector (4, 3)
x3= column vector (7, -3)

Why must x1, x2, and x3 be linearly dependent?

x1 and x2 span R^2.
The basis are these two columns vectors: (3/2, -1/2), (-2, 1)

Since x1 and x2 form the basis, x3 can be written as a linear combination of these vectors.

Is that it? or correct?
 
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  • #2
Dustinsfl said:
x1= column vector (2, 1)
x2= column vector (4, 3)
x3= column vector (7, -3)

Why must x1, x2, and x3 be linearly dependent?

How to answer that question depends on what you have learned. What is the dimension of R2?
x1 and x2 span R^2.
The basis are these two columns vectors: (3/2, -1/2), (-2, 1)

There is no such thing as the basis for R2. Any two linearly independent vectors in R2 are a basis.
Since x1 and x2 form the basis, x3 can be written as a linear combination of these vectors.

Is that it? or correct?

You could just demonstrate x3 = cx1 + dx2; that would surely settle it.
 
  • #3
New question:
x1=(3, -2, 4)
x2=(3, -1, 4)
x3=(-6, 4, -8)

What is the dimension of span (x1, x2, and x3)

The book says 1; however, shouldn't the dimension be 3? I see that these 3 vectors are all the same times a constant but there are coordinates.
 
  • #4
Dustinsfl said:
New question:
x1=(3, -2, 4)
x2=(3, -1, 4)
x3=(-6, 4, -8)

What is the dimension of span (x1, x2, and x3)

The book says 1; however, shouldn't the dimension be 3? I see that these 3 vectors are all the same times a constant but there are coordinates.

If they are supposed to be a constant times each other you have mistyped something. But assuming that, what is the definition of dimension that you are using? You have to apply that.
 

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